Problem: $\begin{cases}d(1)=\dfrac{1}{12}\\\\ d(n)=d(n-1)\cdot (-6) \end{cases}$ What is the $4^{\text{th}}$ term in the sequence?
Answer: This is a recursive formula. It tells us that the first term is $\dfrac{1}{12}$ and that the common ratio is $-6$. $\begin{aligned} {d(1)}&=\dfrac{1}{12} \\\\ {d(2)}&={d(1)}\cdot (-6)=-\dfrac{1}{2} \\\\ {d(3)}&={d(2)}\cdot (-6)=3 \\\\ {d(4)}&={d(3)}\cdot (-6)=-18 \end{aligned}$ The $4^{\text{th}}$ term is $-18$.